• LOJ #6278. 数列分块入门 2


    LOJ #6278. 数列分块入门 2

    思路&&代码

    区间修改+询问区间内小于某个值x的元素个数

    还是先分块,分成(sqrt{n})个块,分的时候,每个块用一个(vector)来维护块内的值

    每个块内排序,保证块内是有序的(便于用(lower\_bound)统计答案),因为分好了这个块以后如果整体修改不会对块内元素造成影响,如果不是整个块修改的话,可以修改之后再顺便排序,覆盖之前这个块的值

    统计答案的时候整个块用(lower\_bound),局部直接暴力枚举,还是(O(nsqrt{n}))

    #include <cmath>
    #include <cstdio>
    #include <vector>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    const int A = 1e5 + 11;
    
    inline int read() {
    	char c = getchar(); int x = 0, f = 1;
    	for ( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    	for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    	return x * f;
    }
    
    vector <int> v[500];
    
    int n, m, a[A], L[A], R[A], add[A], pos[A];
    
    void reset(int x) {
    	v[x].clear();
    	for(int i = L[x]; i <= R[x]; i++) v[x].push_back(a[i]);
    	sort(v[x].begin(), v[x].end());
    }
    
    void change(int l, int r, int c) {
    	int p = pos[l], q = pos[r];
    	if(p == q) {
    		for(int i = l; i <= r; i++) a[i] += c;
    		reset(p); return;
    	}
    	for(int i = l; i <= R[p]; i++) a[i] += c; reset(p);
    	for(int i = L[q]; i <= r; i++) a[i] += c; reset(q);
    	for(int i = p + 1; i < q; i++) add[i] += c;
    	
    }
    
    int ask(int l, int r, int c, int ans = 0) {
    	int p = pos[l], q = pos[r];
    	if(p == q) {
    		for(int i = l; i <= r; i++) if(a[i] + add[p] < c) ans++;
    		return ans;
    	}
    	for(int i = l; i <= R[p]; i++) if(a[i] + add[p] < c) ans++;
    	for(int i = L[q]; i <= r; i++) if(a[i] + add[q] < c) ans++;
    	for(int i = p + 1; i < q; i++) ans += lower_bound(v[i].begin(), v[i].end(), c - add[i]) - v[i].begin();
    	return ans;
    }
    
    int main() {
    	n = read();
    	for(int i = 1; i <= n; i++) a[i] = read();
    	int t = sqrt(n);
    	for(int i = 1; i <= t; i++) {
    		L[i] = sqrt(n) * (i - 1) + 1;
    		R[i] = sqrt(n) * i;
    	}
    	if(R[t] < n) t++, L[t] = R[t - 1] + 1, R[t] = n;
    	for(int i = 1; i <= t; i++) {
    		for(int j = L[i]; j <= R[i]; j++) {
    			pos[j] = i;
    			v[pos[j]].push_back(a[j]);
    		}
    	}
    	for(int i = 1; i <= t; i++) sort(v[i].begin(), v[i].end());
    	m = n;
    	while(m--) {
    		int opt = read(), l = read(), r = read(), c = read();
    		if(opt == 0) change(l, r, c);
    		else cout << ask(l, r, c * c) << '
    ';
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/loceaner/p/12210381.html
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